Does there exists $g\colon (0,1)^2 \to \mathbb{R}$ such that $\int_0^1 g(x,y) dy = 1$ and there exists $f$ satisfying $f(y) = \int_0^1 g(x,y)f(x) dx$

Question

Does there exists a function $g\colon (0,1) \times (0,1) \to \mathbb{R}$ such that

• $g$ is non-negative a.e.,
• $g = 0$ on some set of positive measure,
• $\int_0^1 g(x,y) dy = 1$ for $x \in (0,1)$,
• there exists a function $f\colon (0,1) \to \mathbb{R}$ which is strictly positive a.e. and satisfies $$ f(y) = \int_0^1 g(x,y) f(x) dx, \qquad y \in (0,1)? $$

Is it possible to give an example of such $g$?

Answer 1

If you define $g\ \colon\ (x,y) \longmapsto\begin{cases} 0&\text{if $x\le 0.5\ $ and $y<0.5$}\\ 0&\text{if $x>0.5\ $ and $y\ge 0.5$}\\ 2&\text{otherwise} \end{cases}$

You take $f=1$, $g$ and $f$ satisfy your conditions.

Answer 2

If you are interested in an approach via the theory of linear opreators: Denote $X=[0,1]$, let $g(x,y) \in L^2(X\times X)$ such that $\|g\|_2^2 = \int_{X\times X} |g(x,y)|^2 = 1$ and also $g(x,y) = \overline{g(y,x)}$, We then define a linear map $G\colon L^2 (X) \rightarrow L^2(X) $ by: $$ Gf(y) = \int_X g(x,y)f(x) \mathrm{d}x$$

$G$ is hilbert-schmidt integral operator, and from the theory of such operators we know that: $\|G\|_{\mathrm{op}}= \|g\|_2=1$ and that $G$ is compact.Since $g$ has this certain symmetry in respect to coordinate exchange - we realize $G$ is self-adjoint. Since $G$ is compact and self-adjoint; there exists an eigenvalue $\lambda^*$ such that $|\lambda^*| = \|G\|_{\mathrm{op}}=1$. As a final note, $\lambda^* = \pm 1$ since $G$ is self-adjoint and so every eigenvalue is real.

That being said; Any function $g$ with the above conditions will give you an eigenvalue $\pm 1$ with a corresponding eigenvector (eigen-function) $f$, and after a few attempts you should come up with the eigenvalue $1$ and the function $f$:

$$Gf(y) = f(y) = \int_X g(x,y)f(x)\mathrm{d}x$$